Leetcode 905. Sort Array By Parity, Java解法_905. sort array by parity java-优快云博客

本文链接：https://blog.youkuaiyun.com/YuanTheCoder/article/details/95595262

博客介绍了对非负整数数组进行排序，使偶数在前奇数在后的两种方法。方法一是借鉴快速排序思想的单循环双指针法，时间复杂度O(N)，空间复杂度O(1)；方法二是先放偶数再放奇数的两次遍历法，时间复杂度O(N)，空间复杂度O(N)。

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Given an array A of non-negative integers, return an array consisting of all the even elements of A, followed by all the odd elements of A.

You may return any answer array that satisfies this condition.

Example 1:

Input: [3,1,2,4]
Output: [2,4,3,1]
The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.

Note:

1 <= A.length <= 5000
0 <= A[i] <= 5000

Approach #1 我的解法，one loop, double pointer

class Solution {
    public int[] sortArrayByParity(int[] A) {
        int len = A.length;
        int j = 0;
        int k = len-1;
        while(j<k){
            if(A[j]%2==0){
                j+=1; 
            } 
            else{
                int temp = A[k];
                A[k] = A[j];
                A[j] = temp;
                k-=1;
            }
        }
        return A;
    }
}

这个方法其实借鉴了快速排序算法的思想。

Complexity Analysis

Time Complexity: O(N), where N is the length of A. Each step of the while loop makes j-i decrease by at least one. (Note that while quicksort is O(Nlog N)normally, this is O(N)O(N) because we only need one pass to sort the elements.)
Space Complexity: O(1) in additional space complexity.

Approach #2 Two Pass

class Solution {
    public int[] sortArrayByParity(int[] A) {
        int[] ans = new int[A.length];
        int t = 0;

        for (int i = 0; i < A.length; ++i)
            if (A[i] % 2 == 0)
                ans[t++] = A[i];

        for (int i = 0; i < A.length; ++i)
            if (A[i] % 2 == 1)
                ans[t++] = A[i];

        return ans;
    }
}

先把所有even number放进去，再把所有odd number 放进去。

Complexity Analysis